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Question: Answered & Verified by Expert
3-Phenylpropene on reaction with $\mathrm{HBr}$ gives (as a major product)
ChemistryHaloalkanes and HaloarenesAIIMSAIIMS 2005
Options:
  • A $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3$
  • B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_2 \mathrm{CH}_3$
  • C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$
  • D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}=\mathrm{CH}_2$.
Solution:
1620 Upvotes Verified Answer
The correct answer is: $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_2 \mathrm{CH}_3$
According to Markownikoff's rule, the negative part of the unsymmetrical reagent adds to less hydrogenated (more substituted) carbon atom of the double bond.

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